# 3106. Lexicographically Smallest String After Operations With Constraint

## Description

You are given a string s and an integer k.

Define a function distance(s1, s2) between two strings s1 and s2 of the same length n as:

• The sum of the minimum distance between s1[i] and s2[i] when the characters from 'a' to 'z' are placed in a cyclic order, for all i in the range [0, n - 1].

For example, distance("ab", "cd") == 4, and distance("a", "z") == 1.

You can change any letter of s to any other lowercase English letter, any number of times.

Return a string denoting the lexicographically smallest string t you can get after some changes, such that distance(s, t) <= k.

Example 1:

Input: s = "zbbz", k = 3

Output: "aaaz"

Explanation:

Change s to "aaaz". The distance between "zbbz" and "aaaz" is equal to k = 3.

Example 2:

Input: s = "xaxcd", k = 4

Output: "aawcd"

Explanation:

The distance between "xaxcd" and "aawcd" is equal to k = 4.

Example 3:

Input: s = "lol", k = 0

Output: "lol"

Explanation:

It's impossible to change any character as k = 0.

Constraints:

• 1 <= s.length <= 100
• 0 <= k <= 2000
• s consists only of lowercase English letters.

## Solutions

### Solution 1: Enumeration

We can traverse each position of the string $s$. For each position, we enumerate all characters less than the current character, calculate the cost $d$ to change to this character. If $d \leq k$, we change the current character to this character, subtract $d$ from $k$, end the enumeration, and continue to the next position.

After the traversal, we get a string that meets the conditions.

The time complexity is $O(n \times |\Sigma|)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$, and $|\Sigma|$ is the size of the character set. In this problem, $|\Sigma| \leq 26$.

• class Solution {
public String getSmallestString(String s, int k) {
char[] cs = s.toCharArray();
for (int i = 0; i < cs.length; ++i) {
char c1 = cs[i];
for (char c2 = 'a'; c2 < c1; ++c2) {
int d = Math.min(c1 - c2, 26 - c1 + c2);
if (d <= k) {
cs[i] = c2;
k -= d;
break;
}
}
}
return new String(cs);
}
}

• class Solution {
public:
string getSmallestString(string s, int k) {
for (int i = 0; i < s.size(); ++i) {
char c1 = s[i];
for (char c2 = 'a'; c2 < c1; ++c2) {
int d = min(c1 - c2, 26 - c1 + c2);
if (d <= k) {
s[i] = c2;
k -= d;
break;
}
}
}
return s;
}
};

• class Solution:
def getSmallestString(self, s: str, k: int) -> str:
cs = list(s)
for i, c1 in enumerate(s):
for c2 in ascii_lowercase:
if c2 >= c1:
break
d = min(ord(c1) - ord(c2), 26 - ord(c1) + ord(c2))
if d <= k:
cs[i] = c2
k -= d
break
return "".join(cs)


• func getSmallestString(s string, k int) string {
cs := []byte(s)
for i, c1 := range cs {
for c2 := byte('a'); c2 < c1; c2++ {
d := int(min(c1-c2, 26-c1+c2))
if d <= k {
cs[i] = c2
k -= d
break
}
}
}
return string(cs)
}

• function getSmallestString(s: string, k: number): string {
const cs: string[] = s.split('');
for (let i = 0; i < s.length; ++i) {
for (let j = 97; j < s[i].charCodeAt(0); ++j) {
const d = Math.min(s[i].charCodeAt(0) - j, 26 - s[i].charCodeAt(0) + j);
if (d <= k) {
cs[i] = String.fromCharCode(j);
k -= d;
break;
}
}
}
return cs.join('');
}